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# The gas mileage for Peter’s car is $21$ miles per gallon when the car travels at an average speed of $50$ miles per hour. The car's gas tank has $17$ gallons of gas at the beginning of a trip. If Peter’s car travels at an average speed of $50$ miles per hour, which of the following functions $f$ models the number of gallons of gas remaining in the tank $t$ hours after the trip begins?$A)f(t) = 17 - \dfrac{{21}}{{50t}}$$B)f(t) = 17 - \dfrac{{50}}{{21t}}$$C)f(t) = \dfrac{{17 - 21t}}{{50}}$$D)f(t) = \dfrac{{17 - 50t}}{{21}}$ Verified
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Hint: First we have to define what the terms we need to solve the problem are.
As per the given problem we need to find gallons of gas after the trip.
Since we first analyzed what the given terms are as follows; the gas mileage is twenty-one per gallon and with respect to Peter's car; also, the average speed is fifty miles per hour.
Finally, the car gas tank has seventeen gallons of gas at the beginning.
Now we simply solve this problem by checking the number of options which means first take the option A as follows; $A)f(t) = 17 - \dfrac{{21}}{{50t}}$ since t is the trip begins after $t$ hours but as per given the average speed if $50$ miles per hour so there is no possible of t to be the multiply of fifty and hence it is wrong
Let us now take the option C and D which are $C)f(t) = \dfrac{{17 - 21t}}{{50}}$,$D)f(t) = \dfrac{{17 - 50t}}{{21}}$
Since as we see the options contains the seventeen minus of the $t$-hours in numerator
Therefore the only option left and it is also the correct option which is $B)f(t) = 17 - \dfrac{{50}}{{21t}}$ as we see correctly t-hours is times (multiplied by 21) and seventeen overall minus the average and $t$-hours
$f(t)$ may be a conversion of the polynomial factors.